Search found 1175 matches
- Thu Oct 02, 2014 7:42 pm
- Forum: Number Theory
- Topic: $a\equiv b$ whenever $F(a)\equiv F(b) (mod p)$
- Replies: 8
- Views: 5930
Re: $a\equiv b$ whenever $F(a)\equiv F(b) (mod p)$
Notice that \[F(n) = \sum_{k=1}^{p-1} kn^{k-1} = \sum_{k=1}^{p-1} \dfrac{\text{d}}{\text{d}n} n^k=\dfrac{\text d}{\text d n}\sum_{k=1}^{p-1} n^k=\dfrac{\text d}{\text d n}\left(\dfrac{n^p-n}{n-1}\right)=\dfrac{pn^{p-1}-1}{n-1}-\dfrac{n^p-n}{(n-1)^2}.\] There are major controversies about using calc...
- Wed Oct 01, 2014 9:57 pm
- Forum: Number Theory
- Topic: Dat cool problem
- Replies: 3
- Views: 3382
Re: Dat cool problem
Hint 1 (easy):
${\color{White} {\text{For the minimum such series, the averages are } 1^2, 2^2, \cdots (n-1)^2}}$
Hint 2:
${\color{White} {\text{Split the rangesto the limit, and prove no further improvement is possible }}}$
${\color{White} {\text{For the minimum such series, the averages are } 1^2, 2^2, \cdots (n-1)^2}}$
Hint 2:
${\color{White} {\text{Split the rangesto the limit, and prove no further improvement is possible }}}$
- Wed Oct 01, 2014 7:48 pm
- Forum: Combinatorics
- Topic: Turkey TST 2014
- Replies: 7
- Views: 77598
Re: Turkey TST 2014
Hint:
${\color{White} {\text{Minimize the intersections, and you've minimized the number of worms} } }$
${\color{White} {\text{Minimize the intersections, and you've minimized the number of worms} } }$
- Wed Oct 01, 2014 7:21 pm
- Forum: Secondary Level
- Topic: Directed Angle
- Replies: 3
- Views: 4042
Re: Directed Angle
To add just a little to what Adib said-
Anti-clockwise = positive, Clockwise = negative is the general convention while using the +/- sign.
I was going to post a link here but didn't find anything well-explained enough in google (at least in the first few links).
Anti-clockwise = positive, Clockwise = negative is the general convention while using the +/- sign.
I was going to post a link here but didn't find anything well-explained enough in google (at least in the first few links).
- Wed Oct 01, 2014 7:14 pm
- Forum: Number Theory
- Topic: Quadratic Residues Modulo Prime
- Replies: 6
- Views: 4422
Re: Quadratic Residues Modulo Prime
Note: If we find one quadratic non-residue $\equiv a^2+b^2 \pmod p$, we are done as we can multiply both sides with $z^2$ where $z$ is a primitive root and cycle through the other quadratic non-residues. Now, assume the contrary, it yields $\text{every quadratic residue}+1 = \text{another quadratic ...
- Tue Sep 30, 2014 8:22 pm
- Forum: Combinatorics
- Topic: USAMO 2007
- Replies: 2
- Views: 3223
Re: USAMO 2007
I think $A \cup x =C$ should be $A=C \cup x$.
Other than that, it seems fine.
Other than that, it seems fine.
- Sat Sep 27, 2014 6:29 pm
- Forum: News / Announcements
- Topic: Reduce the usage of BBCodes
- Replies: 2
- Views: 10829
Re: Reduce the usage of BBCodes
Nice initiative- I was thinking of asking everyone having bbcode in their signature to change it if possible (a little scraping odd-job)
But you know, even a little time stings (especially now)
But you know, even a little time stings (especially now)
- Fri Sep 26, 2014 9:28 am
- Forum: Algebra
- Topic: Another Tricky FE
- Replies: 3
- Views: 3381
Re: Another Tricky FE
If you're using $\dfrac{f(x)}x -x^2 =$ constant, you have to prove $f(0)=0$ separately as $g(x) = \dfrac{f(x)}x -x^2$ is not defined at $x=0$.
$x=\frac{a+b}2, y=\frac{a-b}2$ can also be handy whenever the terms are sums and products of $x+y \text{ or } x-y$
$x=\frac{a+b}2, y=\frac{a-b}2$ can also be handy whenever the terms are sums and products of $x+y \text{ or } x-y$
Re: Tricky FE
\[f(n+1)=f(n) + \frac 1{n+2} - \frac 1{n+3}\]
Telescope.
Telescope.
- Mon Sep 22, 2014 7:28 pm
- Forum: Number Theory
- Topic: Magic-Square preserving functional equation(Self-made)
- Replies: 4
- Views: 3788
Re: Magic-Square preserving functional equation(Self-made)
Using this generalized latin square \[\begin{matrix}
{n-1} & {n+1} & n \\
{n+1} & n & {n-1} \\
n & {n-1} & {n+1}
\end{matrix}\]
We get $f(n) = \frac{f(n+1)+f(n-1)}2$ which implies $f$ is linear, i.e $f(x)$ has the form $ax+b$.
It can be easily verified that this $f$ satisfies the conditions.
{n-1} & {n+1} & n \\
{n+1} & n & {n-1} \\
n & {n-1} & {n+1}
\end{matrix}\]
We get $f(n) = \frac{f(n+1)+f(n-1)}2$ which implies $f$ is linear, i.e $f(x)$ has the form $ax+b$.
It can be easily verified that this $f$ satisfies the conditions.